In the figure,$AB$ is a diameter of the circle,and $CD$ is a chord equal to the radius of the circle. $AC$ and $BD$ when extended intersect at a point $E$. Prove that $\angle AEB = 60^{\circ}$.

(N/A) Join $OC$,$OD$,and $BC$.
Since $CD$ is equal to the radius of the circle $(OC = OD = CD)$,triangle $ODC$ is an equilateral triangle.
Therefore,$\angle COD = 60^{\circ}$.
Now,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Thus,$\angle CBD = \frac{1}{2} \angle COD = \frac{1}{2} \times 60^{\circ} = 30^{\circ}$.
Since $AB$ is the diameter,the angle in a semicircle is a right angle,so $\angle ACB = 90^{\circ}$.
Since $ACE$ is a straight line,$\angle BCE = 180^{\circ} - \angle ACB = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
In $\triangle BCE$,the sum of angles is $180^{\circ}$.
Therefore,$\angle CEB + \angle BCE + \angle CBE = 180^{\circ}$.
$\angle CEB + 90^{\circ} + 30^{\circ} = 180^{\circ}$.
$\angle CEB = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
Hence,$\angle AEB = 60^{\circ}$.